TITLE: Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.
QUESTION [19 upvotes]: Wolfram$\alpha$ says that we have the following identity
$$
\left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}}
$$
but, how does one prove such identity?

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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$$\pars{\pi \over 2}^{2}
     = 1 + 2\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}:\ {\Large ?}
\tag{1}
$$

\begin{align}
&\color{#00f}{\large\sum_{k=1}^{\infty}{\pars{2k - 1}\zeta\pars{2k} \over 2^{2k}}}=
\sum_{k=1}^{\infty}{2k - 1 \over 2^{2k}}\sum_{\ell = 1}^{\infty}{1 \over \ell^{2k}}
=\sum_{\ell = 1}^{\infty}\sum_{k=1}^{\infty}\pars{2k - 1}\pars{1 \over 2\ell}^{2k}
\\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2}
\totald{}{x}\sum_{k=1}^{\infty}x^{2k - 1}}_{x = 1/\pars{2\ell}}
=\sum_{\ell = 1}^{\infty}\braces{x^{2}\,
\totald{}{x}\bracks{x \over 1 - x^{2}}}_{x = 1/\pars{2\ell}}
\\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{x^{2}\,
{1 + x^{2} \over \pars{1 - x^{2}}^{2}}}_{x = 1/\pars{2\ell}}
=\sum_{\ell = 1}^{\infty}\bracks{
{1/x^{2} + 1 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}}
\\[3mm]&=\sum_{\ell = 1}^{\infty}\bracks{
{1 \over 1/x^{2} - 1} + {2 \over \pars{1/x^{2} - 1}^{2}}}_{x = 1/\pars{2\ell}}
=\left.\pars{2\,\totald{}{\mu} + 1}\sum_{\ell = 1}^{\infty}{1 \over 4\ell^{2} - \mu}
\right\vert_{\mu = 1}
\\[3mm]&=\pars{2\,\totald{}{\mu} + 1}\bracks{%
{2 - \root{\mu}\pi\cot\pars{\root{\mu}\pi/2}} \over 4\mu}_{\mu = 1}
=\color{#00f}{\large\half\bracks{\pars{\pi \over 2}^{2} - 1}}\tag{2}
\end{align}

Replace $\pars{2}$ in $\pars{1}$.